How to visualize this statement regarding Conjugate Symmetry

Question

A property of real signals states that if $x(t)$ is real then the Fourier series coefficient (frequency spectrum) is given by : $$ c_{k}=c_{-k}^{*} $$ In polar form, this can be expressed as : $$ |c_{k}|=|c_{-k}|\qquad\text{and}\qquad\arg c_{k}=-\arg c_{-k} $$ It appears here that the magnitude spectrum obeys even symmetry and the phase spectrum obeys odd symmetry. It is said in my textbook :

1. Due to symmetry in the frequency spectra of real signals, we typically ignore negative frequencies when dealing with such signals.
2. In the case of signals that are complex but not real, frequency spectra do not possess the above symmetry, and negative frequencies become important.

For $(1)$, how can I visualize this to see why the negative frequency components do not matter anymore$?$, does this mean the Fourier series representation of $x(t)$ becomes : $$ x(t)=\sum_{k=0}^{\infty}c_{k}e^{jk\omega_{0} t}\;??? $$ For $(2)$, is this statement true because the complex exponentials in the frequency domain will scale a given signal to a decaying or growing signal?

Answer 1

For $(1)$, how can I visualize this to see why the negative frequency components do not matter anymore$?$, does this mean the Fourier series representation of $x(t)$ becomes : $$ x(t)=\sum_{k=0}^{\infty}c_{k}e^{jk\omega_{0} t}\;??? $$

No, the Fourier series is still what it was before this conjugate symmetry nonsense reared its ugly head. But what happens is that

\begin{align} x(t) &= \sum_{k=-\infty}^{\infty}c_{k}e^{jk\omega_{0} t} & {\scriptstyle\text{original definition}}\\ &= \sum_{k=-\infty}^{-1}c_{k}e^{jk\omega_{0} t} + c_0 + \sum_{k=1}^{\infty}c_{k}e^{jk\omega_{0} t}& {\scriptstyle\text{breaking up the sum}}\\ &= \sum_{k=1}^{\infty}c_{-k}e^{-jk\omega_{0} t} + c_0 + \sum_{k=1}^{\infty}c_{k}e^{jk\omega_{0} t}& {\scriptstyle\text{changing indices}}\\ &= c_0 + \sum_{k=1}^{\infty}c_{k}e^{jk\omega_{0} t} + c_{-k}e^{-jk\omega_{0} t} & {\scriptstyle\text{combining sums}}\\ &= c_0 + \sum_{k=1}^{\infty}c_{k}e^{jk\omega_{0} t} + c_{k}^*e^{-jk\omega_{0} t} & {\scriptstyle c_k = c_{-k}^* \implies c_{-k} = c_k^*}\\ &= c_0 + \sum_{k=1}^{\infty}c_{k}e^{jk\omega_{0} t} + \left(c_{k}e^{jk\omega_{0} t}\right)^* & {\scriptstyle a^*b^* = (ab)^*}\\ &= c_0 + 2\sum_{k=1}^{\infty}\operatorname{Re}\left(c_{k}e^{jk\omega_{0} t}\right) \end{align} showing that one needs to compute or know the $c_k$ values for nonnegative $k$ only; the values of $c_k$ for $k < 0$ are not needed. Anal-retentives who believe in crossing their eyes and dotting their teas can always determine the values of the $c_{-k}$ by taking the complex conjugates of the known $c_k$'s.

Answer 2

No, you have a misunderstanding.

The fact that a real-valued signal's Fourier series coefficients $c_k$ are conjugate-symmetric, does not mean that the negative frequencies are ignored. They are still as important as the positive frequencies, yet the symmetry in the coefficients lets you compute only one half of the coefficients, and deduce the remaning half by symmetry.

Note that, you can compute either the positive half or the negative half, and immediately find the remaining half via the conjugate-symmetry condition.

Hence, the negative frequencies in the complex Fourier series representation of a real-valued signal are not ignored and still utilised as much as the positive frequencies.